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\begin{document}
\begin{center}{\Large Basic Algebra 1\\Solutions
of Assignment 6}\end{center}
\noindent{\bf Solution of the problem 1.} Starting with prime number
3 and looking for a root, we see that $f(x)=x^2+1$ has no zero in
$\mathbb{Z}_3$, hence it is irreducible in $\mathbb{Z}_3[x]$. Next
consider 5. This case is actually different. In fact in
$\mathbb{Z}_5[x]$ we have the factorization $f(x)=(x+2)(x+3)$.
Continuing this way, we find that $f(x)$ is irreducible in
$\mathbb{Z}_p[x]$ for $p=3, 7, 11, 19, 23$ and is reducible for
$p=5, 13, 17$.
Looking for a general pattern, first note that each of the primes 5,
13 and 17 is of the form $4k+1$, and on the contrary, none of the
primes 3, 7, 11, 19 and 23 is in that form. Secondly, observe that
$$5=2^2+1^2,~13=3^2+2^2,~17=4^2+1^2,$$
while the primes 3, 7, 11, 19 and 23 don't enjoy such property,
namely they cannot be represented as a sum of two squares. In fact
one has the following beautiful theorem of Fermat:
\medskip\par
{\it An odd prime number $p$ is a sum of two square, i.e.,
$p=a^2+b^2$, if and only if it is of form $4k+1$.}
\medskip\par
\noindent Also look at the solution of the problem 5.
\bigskip\par
\noindent{\bf Solution of the problem 2.} Here is one example:
$f(x)=2x^2+4$. Note that $f(1)=f(2)=f(4)=f(5)=0.$ This does not
contradict the theorem proven in class stating that a polynomial in
$F[x]$ of degree $d$ has at most $d$ roots and the reason is simple:
$\mathbb{Z}_6$ is not a field!
\bigskip\par
\noindent{\bf Solution of the problem 3.} This can be done by a
trial and error search and here is the answer:
$$[x^2+x+1]^{-1}=[x^2].$$
To verify our answer, notice that since $[x^3+x+1]=[0]$, we have
$$[x^2+x+1][x^2]=[x^4+x^3+x^2]=[x(-x-1)+(-x-1)+x^2]=[1]$$ ({\bf N.B.}
$2=0$ and $-1=1$, because we are working in $\mathbb{Z}_2$.) For
another way of looking at this problem, go to the solution of the
next problem.
\bigskip\par
\noindent{\bf Solution of the problem 4.} Here you are:
$$[x]^1=[x],~[x]^2=[x^2],~[x]^3=[x+1],~[x]^4=[x^2+x],$$
$$[x]^5=[x^2+x+1],~[x]^6=[x^2+1],~[x]^7=[1].$$
So, the smallest $j>0$ for which $[x]^j=[1]$ is 7, and therefore
$[x]$ is a generator for the multiplicative group of nonzero
elements of the finite field $\mathbb{Z}_2[x]/(x^3+x+1).$
\medskip\par
Back to the solution of the previous problem, note that
$$[x^2+x+1][x^2]=[x]^5[x]^2=[x]^7=[1]~!$$
\bigskip\par
\noindent{\bf Solution of the problem 5.} Proof by contradiction.
Suppose that $x^2+1$ factors in $\mathbb{Z}_p[x]$. So, it has a
root, $a$ say, in $\mathbb{Z}_p$, i.e., $a^2+1=0$ in $\mathbb{Z}_p$.
This in turn implies that $p\mid a^2+1$ or equivalently
$a^2\equiv-1\pmod p$. Now since $p$ is odd, we can raise both sides
of $a^2\equiv-1\pmod p$ to the power $\displaystyle{\frac{p-1}{2}}$
to get
$$a^{p-1}\equiv(-1)^{\frac{p-1}{2}}\pmod p.$$
Comparing with little Fermat, we infer that
$$1\equiv(-1)^{\frac{p-1}{2}}\pmod p.$$
Since $p>2$, this is impossible unless the last congruence relation
becomes equality and that amounts to $\frac{p-1}{2}=2k$, hence
$p=4k+1$, which is a contradiction!
\bigskip\par
\bigskip\par
\noindent{\bf Solution of the problem 6a.} Let $R=F[X]$ where $F$ is
a field and let $I=F$ be the set of constant polynomials. Then $I$
isn't an ideal because $i \in I \IFF \deg(i)=0 \textrm{~or~} i=0$.
So for any $f\in R$ s.t. $\deg(f)\ge 1$ and $0 \neq i \in I$ we have
$\deg(f*i) = \deg(i)+deg(f) \ge 1 \ifthen f*i \notin I$.
\bigskip\par
\noindent {\bf Solution of the problem 6b.} Let $R=\Z \times \Z$ and
let $I=\{(m,0)|m\in \Z \}$. Then $I$ is an ideal. Let $(a,0), (b,0)
\in I$. Then $(a,0) + (b,0) = (a+b,0)
\in I$, so $I$ is closed under addition and if $(m,n)\in R$ is some arbitrary
element we have that $(m,n)*(a,0) = (ma,0) \in I$ so $I$ is closed
under multiplication by arbitrary elements in $R$. It follows that
$I$ is an ideal.
\bigskip\par
\noindent {\bf Solution of the problem 6c.} Let $I$ be the set of
nilpotent elements of a (commutative) ring $R$ i.e.
$$I= \{a\in R |
\exists~m \in \mathbb{N}~{\rm s.t.}~a^m=0 \}.$$ Then $I$ is an
ideal. Let $a,b \in I$, then $\exists m,n \in \mathbb{N}$ s.t. $a^m
= b^n = 0$. If $r\in R$ is some arbitrary element then $(ra)^m =
r^{m}a^{m} = r^{m}0=0$ (the second equality holds because $R$ is
commutative), so $ra$ is also nilpotent, hence $I$ is closed under
multiplication by arbitrary elements. Now let $N=\max(n,m)$ and
consider $(a+b)^{2N}$. First note that if $i\leq N$ then $2N-i \geq
N$. So by the binomial theorem we have: \begin{eqnarray*}
(a+b)^{2N}&=&\sum_{i=0}^{2N}{{2N}\choose{i}}a^{i}b^{2N-i}\\
&=&\sum_{i=0}^{N}{{2N}\choose{i}}a^{i}\underbrace{b^{2N-i}}_{=0}+
\sum_{i=N+1}^{2N}{{2N}\choose{i}}\underbrace{a^{i}}_{=0}b^{2N-i}\\
&=&0 \end{eqnarray*}
It follows that $a+b$ is also nilpotent, so $I$
is closed under addition, it follows that $I$ is an ideal.
\medskip\par
\noindent{\bf Remark} It is important to notice that the statement
of this problem is false if we remove the commutativity of $R$. Here
is one example. Let $R=M_2[\mathbb{Z}]$ be the ring of all
$2\times2$ integer matrices (with the usual addition and
multiplication), and let
$$A={{0~1}\choose{0~0}},~~~B={{0~0}\choose{1~0}}.$$
One easily sees that
$$A^2=B^2=O_2~({\rm the~zero~matrix~of~size~2},)$$
however, no positive power of
$\displaystyle{A+B={{0~1}\choose{1~0}}}$ is zero! (check this). The
reason behind this is that $M_2[\mathbb{Z}]$ is not a commutative
ring.
\bigskip\par
\bigskip\par
For the next two parts, let $R$ be the ring of functions from $\Z$
to $\mathbb{R}$ with addition $+_{R}$ and multiplication $*_{R}$.
For $f,g \in R$, the addition $f +_{R} g \in R$ is defined as the
mapping:
\[(f +_{R} g) (x) = f(x)+g(x)\]
and the product of two functions $f$ and $g$ is defined as the
mapping:
\[(f*_{R}g) (x) = f(x)g(x).\]
From now on, subscripts for the operation signs will be omitted.
\bigskip\par
\noindent {\bf Solution of the problem 6d.} Let $I$ be the set of
functions $f$ s.t. $f(0)=f(1)$. $I$ is not an ideal. Indeed, let
$f\in I$ be the constant $1$ function i.e. for each $n\in \Z,
f(n)=1$ and let $g$ be the function $g(n)=n$. Then we have that
$(f*g)(0)=f(0)g(0)=1\times0=0$ but $(f*g)(1)=f(1)g(1)=1$. So
$f*g\notin I$ so $I$ is not closed under multiplication by arbitrary
elements of $R$ so it's not an ideal.
\bigskip\par
\noindent {\bf Solution of the problem 6e.} Let $I=\{f \in R |
f(0)=f(1)=0\}$. Let $f,g \in I$ then $(f+g)(1)=f(1)+g(1)=0+0=0$ and
similarly $(f+g)(0)=0$ so $I$ is closed under addition. Now let $h
\in R$ be some arbitrary element. Then we find that
$(h*f)(1)=h(1)f(1)=h(1)\times0=0$ and
$(h*f)(0)=h(0)f(0)=h(0)\times0=0$, so $h*f \in I$. It follows that
$I$ is closed under multiplication by arbitrary elements of $R$. It
follows that $I$ is an ideal.
\bigskip\par
\noindent {\bf Solution of the problem 7.} Let $R$ be the polynomial
ring $F[X]$ with coefficients in a field. Then all of its ideals are
principal.
\medskip\par
\noindent{\bf Remark.} In class the strategy was to show that an
ideal in $\Z$ is generated by its smallest positive element. Recall
that in polynomial rings over fields, the notion of size corresponds
to the degree of a polynomial.
\noindent {\bf Proof:} Let $I\subset F[X]=R$ be an ideal. Then the
set $S=\{n \in \mathbb{N}~:~n = \deg(f),~{\rm fo~some}~f \in I\}
\subset \mathbb{N}$ is nonempty (why?), so it must have a smallest
element. If $0\in S$ then $I$ contains a constant $\ifthen$ $I=R$.
Otherwise let $s>0$ be the smallest element in $S$. Then there is
some $f\in I$ s.t. $\deg(f)=s$, i.e. $f$ is the element of minimal
degree in $I$.
Now we claim that $I=(f)$. On one hand $f \in I \ifthen (f)\subseteq
I$. On the other hand, suppose $I \nsubseteq (f)$, then there is
some $g\in I$ such that $f\nmid g$. So we can apply the division
algorithm to get
\[g = fq + r\] where $r \neq 0$ and $\deg(r)<\deg(f)=s$. But then we
get $r=g-fq \in I$ because of closure of ideals under addition and
multiplication. And we get that $m=\deg(r)\in S$ and $m